Expert: Alon Mandes - 8/13/2009

Question
QUESTION: the constant a is such that integration of (xe^.5x) with limits a to 0 equal to 6.Show that a satisfies the equation x=2+e^(-.5x). Thanks

ANSWER: The integral of xe^(x/2) is 2(x-2)e^(x/2). This can be easily shown
via "Integration By Parts" method :
Let's set x=v & e^(x/2)=du . Thus :
INT of xe^(x/2) = 2xe^(x/2) - INT of 2e^(x/2)
         = 2xe^(x/2)-4e^(x/2)
         = 2(x-2)e^(x/2).
So,
INT from 0 to a yields : 2(a-2)e^(a/2) + 4 .
Hence, our equation for a will become :
6=2(a-2)e^(a/2) + 4
2=2(a-2)e^(a/2)
a-2=e^(-a/2)
a=2+e^(-a/2) .

Alon.



---------- FOLLOW-UP ----------

QUESTION: Can you extend your solutions for INT of xe^(x/2), please? Because I can't get to your next steps which is 2xe^(x/2)- INT of 2e^(x/2). I'm very weak with this topic. Thanks a million

ANSWER: Ok, let's elaborate :

1st :
---
∫αe^(βx) dx = (α/β)e^(βx) . It's a basic rule of integration . ("you are welcome to chick it").

2nd :
---
"Integration by Parts " method consisted of a basic idea or rule, which says :
∫u*v'=u*v-∫u'*v . Or it can be written also as : ∫u*dv=u*v-∫du*v . It may also be written as:
∫f(x)g'(x) dx = f(x)g(x)-∫f'(x)g(x) dx . all these 3 form are the same.
Now,
We are interested in calculating ∫xe^(½x) dx . Do to so, we use the method of integration by parts.
Let's perform it :
∫  x  *   e^(½x)  dx
∫  u  *     v'    dx
Here we chose u=x & e^(½x)=v' . Therefore u'=1 & v=2e^(½x) .
Let's follow the rule : ∫u*v'=u*v-∫u'*v .
∫  u   *   v'        =   u  *   v         -   ∫   u'   *  v
∫  x   *   e^(½x)    =   x  *   2e^(½x)   -   ∫   1    *  2e^(½x)
I hope tell now it's clear .
Now, we know that ∫2e^(½x) dx = 4e^(½x). ("please verify it"), So we have :
∫xe^(½x) = 2xe^(½x) - 4e^(½x). We solved the integral.
All that left to do now is to substitute the limits of the integration .

Alon.




---------- FOLLOW-UP ----------

QUESTION: Thanks it really helpful I kinda get it a bit already but I got another question which is why we need to do the first integration which we get 2xe^.5x then do the second integration which is integration by parts by making use of 2xe^.5x? What I mean is why we need to do two times of integration? Many thanks

Answer
The integration of exponential functions :
1. Since (e^x)'=e^x , we conclude that ∫e^x dx = e^x .
2. Since (e^[ax])'=ae^ax , we conclude that ∫e^[ax] dx = (1/a)e^[ax] .
But the problem rises when we want to integrate xe^x or xe^(ax) . It's not an immediate
integration such as the forms, but another technique must be applied . This method will be
integration by parts. That's the reason why we performed twice .

Alon.